use reduction of order to solve the problem...

**slapmaxwell1** · March 25th 2011, 05:40 PM

y''-4y=2; y1=e^(-2x)

i dont see how my text got yp=1/2

ok so this is what i have so far....

let y2=ue^(-2x)

$y^{'}=u^{'}e^{-2x}-2ue^{-2x}$

$y^{''}=u^{''}e^{-2x}-4u^{'}e^{-2x}+4ue^{-2x}$

so $y^{''}-4y = u^{''}e^{-2x}-4u^{'}e^{-2x}$

$e^{-2x}[u^{''}-4u^{'}]=0 ; w=u^{'}$

$\int (e^{-4x}w)^{'}=\int 0$

$w = \frac{c_{1}}{e^{-4x}}$

$\int u^{'}=\int \frac{c_{1}}{e^{4x}}; u=c_{2}e^{4x}$

so $y_{2}=c_{2}e^{4x}e^{-2x}\rightarrow y_{2}=c_{2}e^{2x}$

$y=c_{1}e^{-2x}+c_{2}e^{2x}$

i have the yc part i just dont know how to get the yp..what am i missing or not getting??

**derdack** · March 27th 2011, 02:58 AM

x''(t)-4x(t)=2. First find the homogeneous x''(t)-4x(t)=0, characteristic equation m^2-4=0, m1/2=+-2, xh=C1*e^(m1*t)+C2*e^(m2*t)=C1e^(2*t)+C2*e^(-2*t). As we see that a non homogeneous part in start equation is 2, xp must be constant. x=xh+xp. Substitute this in start equation (C1*e^2t+C2*e(-2t)+xp)''-4(C1*e^2t+C2*e(-2t)+xp)=2, xp''=0, -4xp=2, xp=-1/2.

Kind regards,

**HallsofIvy** · March 27th 2011, 03:31 AM

derdack, yes, the problem can be solved that way, but I don't believe it is a god idea to advise a person to ignore the teacher's or text's instructions. Slapmaxwell1 says he was instructed to use "reduction of order". It is not at all uncommon to ask a person to solve a simple problem using a more complicated technique in order to practice that technique.

Slapmaxwell1, yes, letting $y= ue^{2x}$ reduces the left side of the equation to $e^{-2x}(u''- 4u')$ but you appear to have left out the right side of the equation. You have $e^{-2x}(u''- 4u')= 2$ so that $u''- 4u'= 2e^{-2x}$ and, letting w= u', $w'- 4w= 2e^{-2x}$ .

That is now a first order equation linear equation. It has $e^{-4x}$ as integrating factor:
$e^{-4x}w'- 4e^{-4x}w= (e^{-4x}w)'= 2e^{-4x}$

**derdack** · March 27th 2011, 06:12 AM

You are right, but it is completly corect that we have other ways for solving which is important for learner. (John - Beautiful mind) .

**topsquark** · March 27th 2011, 07:25 AM

Originally Posted by derdack

You are right, but it is completly corect that we have other ways for solving which is important for learner. (John - Beautiful mind) .

True, but I was taught reduction of order before your method. The different methods come in their own itme.

-Dan

**derdack** · March 27th 2011, 01:12 PM

Yes. Both of us are right...

**slapmaxwell1** · March 27th 2011, 01:16 PM

i am starting to see why you guys were so pissed that i wasnt using latex...I apologize. learning the commands in beginning is a bit troublesome and take a lil extra time, but its totally worth the effort.

**slapmaxwell1** · March 27th 2011, 01:52 PM

ok ok i see it now.

if i let $y_{p}=A$ then $y'=0 , y''=0$

ok so $0-4A=2$ so $A=\frac{-1}{2}$

there for $y_{p}=\frac{-1}{2}$ so $y=y_{c}+y_{p}$ because i found Yc I would just put it all together for my final answer. thanks again!

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